$\int \sin^3(x)\cos(x)\,dx\,= $ $+~C$
Answer: If we let $ {u=\sin(x)}$, then $du=\cos(x) \, dx}$. Substituting gives us: $\begin{aligned} \int {\sin}^3{(x)}\cos(x)\,dx} &= \int u^3\,D {du}\\\\\\\end{aligned}$ We recognize this antiderivative. $\begin{aligned}\phantom{\int\dfrac{e^x}{1+e^{2x}}dx}&= \int u^3\, du\\\\\\ &=\dfrac{1}{4}u^4+C\\\\\\\end{aligned}$ We can now substitute back to find the antiderivative in terms of $x$. ∫ e x 1 + e 2 x d x = 1 4 u 4 + C = 1 4 sin 4 ( x ) + C \begin{aligned}\phantom{\int\dfrac{e^x}{1+e^{2x}}dx~}&=\dfrac{1}{4}u^4+C\\\\\\\ &=\dfrac14\sin^4(x)+C\end{aligned} The answer: $\int \sin^3(x)\cos(x)\,dx\,= \dfrac14\sin^4(x)+C$